Chapter 11 polar coordinate - National Tsing Hua...

Chapter 11 polar coordinate

Speaker: Lung-Sheng Chien

Book: Lloyd N. Trefethen, Spectral Methods in MATLAB

Discretization on unit diskConsider eigen-value problem on unit disk

2u uλ∆ = − with boundary condition ( )1, 0u r θ= =

We adopt polar coordinate cossin

x ry r

θθ

=⎧⎨ =⎩

, then

22

1 1rr ru u u u u

r r θθ λ∆ = + + = − ( ]0,1r∈ [ ]0,2θ π∈, on and

, and non-periodic Chebyshev grid in rUsually we take periodic Fourier grid in θ

[ ]0,1r∈1

Chebyshev grid in [ ]1,1x∈ −

12

xr +=

Chebyshev grid in [ ]0,1r∈

Observation: nodes are clustered near origin 0r = , for time evolution problem,

we need smaller time-step to maintain numerical stability.

[ ]1,1r∈ −2

( )( )

cos cos

sin sin

x r r

y r r

θ θ π

θ θ π

⎧ = = − +⎪⎨

= = − +⎪⎩( ) ( ), 0,0x y ≠

1 to 2 mapping ( ),r θ

Asymptotic behavior of spectrum of Chebyshev diff. matrix

2NDIn chapter 10, we have showed that spectrum of Chebyshev differential matrix

(second order) approximates

with eigenmode2

2

4k kπλ = −( ), 1 1, B.C. 1 0xxu u x uλ= − < < ± =

2ND1 4

max 0.048Nλ ≈ −Eigenvalue of is negative (real number) and2

2

4k kπλ = −Large eigenmode of does not approximate to2ND2

Since ppw is too small such that

resolution is not enough

Mode N is spurious and localized near boundaries 1x = ±

Grid distribution

[ ]0,1r∈ [ ]1,1r∈ −1 2

Preliminary: Chebyshev node and diff. matrix [1]

Consider 1N + Chebyshev node on [ ]1,1− , cosjjxNπ⎛ ⎞= ⎜ ⎟

⎝ ⎠for 0,1,2, ,j N=

x0x1x2x3x/ 2NxNx

1− 1

6N =

0 1x =1x2x3 0x =4x5x61 x− =x

Even case:

Uniform division in arc

Odd case:

0 1x =1x2x03x4x51 x− =x

5N =


Given 1N + Chebyshev nodes , cosjjxNπ⎛ ⎞= ⎜ ⎟

⎝ ⎠and corresponding function value jv

We can construct a unique polynomial of degree N , called ( ) ( )0

N

j jj

p x v S x=

=∑

( ) 1 0 j k

k jS x

k j=⎧

= ⎨ ≠⎩is a basis.

( ) ( ) ( )1

0 0

N NN

i j j i ij jj j

p x v S x D v= =

′ = =∑ ∑ where differential matrix ( )NN ijD D is expressed as

2

002 1,

6N ND +=

22 1,6

NNN

ND += − ( )2

,2 1

jNjj

j

xD

x−

=−

for 1,2, , 1j N= −

( )1,

i jN iij

j i j

cDc x x

+−=

− for , , 0,1,2, ,i j i j N≠ = where2 0,1 otherwisei

i Nc

=⎧= ⎨⎩

with identity 0,

NN Nii ij

j j iD D

= ≠

= − ∑

Second derivative matrix is 2N N ND D D= ⋅


Let ( )p x be the unique polynomial of degree with N≤ ( )1 0p ± = and ( )j jp x v=

define 0 j N≤ ≤( )j jw p x′′= and ( )j jz p x′= for

, then impose B.C. We abbreviate 2Nw D v= ⋅ ( )1 0p ± = ,that is, 0 0Nv v= =

In order to keep solvability, we neglect 0 Nw w= ( )2 2 1: 1,1: 1N ND D N N= − −,that is,

0w

1w

2w

1Nw −

Nw

0v

1v

2v

1Nv −

Nv

zero

2ND

neglect

neglect

=

zero

( )1: 1,1: 1N ND D N N= − −Similarly, we also modify differential matrix as

Preliminary: DFT [1]

Given a set of data point { }1 2, , , NNv v v R∈ with 2N m= is even,

2hNπ

=

Then DFT formula for { }jv

( )1

2ˆ expN

k j jj

v v ikxNπ

=

= −∑ for , 1, , 1,k m m m m= − − + −

( )1

1 ˆ exp2

m

j k jk m

v v ikxπ =− +

= ∑ for 1,2, ,j N=

Definition: band-limit interpolant of functionδ − , is periodic sinc function ( )NS x

( ) ( )( ) ( )

( )( )

sin / sin12 2 / tan / 2 2 tan / 2

mikx

Nk m

x h mxhS x P eh x m xπ

π π=−

= =∑

If we write 1

N

j k j kk

v v vδ δ−=

= = ∗∑ , then ( ) ( )1

1 ˆ2

m Nikx

k k N kk m k

p x P e v v S x xπ =− =

= = −∑ ∑

Also derivative is according to ( ) ( ) ( )1

1

N

j j k N j kk

w p x v S x x=

′= = −∑


Direct computation of derivative of ( ) ( )( ) ( )

sin /2 / tan / 2N

x hS x

h xπ

π=

( ) ( )

, we have

( )

( ) ( )1

0 0 mod

1 1 cot , 0 mod2 2

jN j

j NS x jh j N

⎧ =⎪= ⎨ ⎛ ⎞− ≠⎜ ⎟⎪

⎝ ⎠⎩

( ) ( )( )

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )

1 1 1 15 5 5 5

1 1 1 15 5 5 5

1 1 1 1 15 5 5 5 5 5

1 1 1 15 5 5 5

1 1 1 15 5 5 5

0 2 3 4

0 2 3

2 0 2

3 2 0

4 3 2 0

j k

S h S h S h S h

S h S h S h S h

D S x x S h S h S h S h

S h S h S h S h

S h S h S h S h

⎛ ⎞− − − −⎜ ⎟

− − −⎜ ⎟⎜ ⎟

= − = − −⎜ ⎟⎜ ⎟−⎜ ⎟⎜ ⎟⎝ ⎠

Example:

is a Toeplitz matrix.

Second derivative is ( ) ( )( )

( )( ) ( )

2

22

2

1 0 mod 6 3

1, 0 mod

2sin / 2

jN j

j Nh

S xj N

jh

π⎧− − =⎪⎪= ⎨ −⎪− ≠⎪⎩


( ) ( )( )

( )( ) ( )

2

22

2

1 0 mod 6 3

1, 0 mod

2sin / 2

jN j

j Nh

S xj N

jh

π⎧− − =⎪⎪= ⎨ −⎪− ≠⎪⎩

For second derivative operation ( ) ( ) ( ) ( )2 2

1 1,

N N

j j k N j k N kk k

w p x v S x x D j k v= =

′′= = − ≡∑ ∑

second diff. matrix is explicitly defined by using Toeplitz matrix (command in MATLAB)

( ) ( )( )

( )( )

( )( )

( )( )( )

2

22:2

22 22 2

2

2

2

csc 2 / 2 / 2csc / 2 / 2

11 ,1 6 3 2sin 1: 1 / 26 3

csc / 2 / 2csc 2 / 2 / 2

N

N N j k

hh

D S x x toeplitzh N h

hh

h

ππ

⎛ ⎞⎜ ⎟−⎜ ⎟⎜ ⎟ ⎛ ⎞−⎜ ⎟ ⎜ ⎟= − = = − −⎜ ⎟ ⎜ ⎟− − −⎝ ⎠⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟−⎝ ⎠

Fornberg’s idea : extend radius to negative image [1]

[ ]1,1r∈ − [ ]0, 2θ π∈( ), x y unit disk∈ and

5rN = (odd): to avoid singularity of coordinate transformation 0r =1

( ) ( )( )( ), , mod 2u r u rθ θ π π= − +6Nθ = (even): to keep symmetry condition2

θ

r

01 r=

1r

2r

3r

4r

51rNr r− = =

0θ 1θ 2θ 3θ 4θ 5θ 6θ0 π 2Nθ

θ π=

0r =

coordinater θ− x y− coordinate

y1θ2θ

3θ

4θ 5θ

6θ x

In general

θ

r

01 r=

1r

kr

1kr +

1rNr −

1rNr− =

0θ 1θ 2θ mθ 1mθ +

0 π 2Nθθ π=

0r =

r θ− coordinate

2r

1Nθθ −

x

y

1θ

2θ

mθ

1mθ +

x y− coordinate

Nθθ

1Nθθ −

2 1rN k= + 2N mθis odd, and is even, then=

Fornberg’s idea : extend radius to negative image [2]

: non-unifr∆ 2Nθ

πθ∆ =

1 : 1 1i rr i r i N

and

= − ∆ ≤ ≤ −

: 1j j j Nθθ θ= ∆ ≤ ≤( )1rN Nθ−Active variable , total number is

Redundancy in coordinate transformation [1]

( ) ( )( )( ) ( ), , , mod 2 ,r r x yθ θ π π− + ↔2 to 1 mapping

θ

r

01 r=

1r

2r

3r

4r

51rNr r− = =

0θ 1θ 2θ 3θ 4θ 5θ 6θ0 π 2Nθ

θ π=

x

y1θ2θ

3θ

4θ 5θ

6θ

x y− coordinatecoordinater θ−

redundant

0r =

y1θ2θ

3θ

4θ 5θ

6θ x


( ) ( )( )( ) ( ), , , mod 2 ,r r x yθ θ π π− + ↔2 to 1 mapping

x y− coordinatecoordinater θ−y

1θ2θ

3θ

4θ 5θ

6θ

θ

r

01 r=

1r

2r

3r

4r

51rNr r− = =

0θ 1θ 2θ 3θ 4θ 5θ 6θ0 π 2Nθ

θ π=

x

0r =

y1θ2θ

3θ

4θ 5θ

6θ

redundant

x

8.5

2.6180

-0.7236

0.3820

-0.2764

0.5

-10.4721

-1.1708

2

-0.8944

0.6180

-1.1056

2.8944

-2

-0.1708

1.618

-0.8944

1.5279

-1.5279

0.8944

-1.6180

0.1708

2

-2.8944

1.1056

-0.618

0.8944

-2

1.1708

10.4721

-0.5

0.2764

-0.382

0.7236

-2.618

-8.5

rND =

0r

0r

1r

1r

2r

2r

3r

3r

4r

4r

5r

5r

Redundancy in coordinate transformation [3]2 1 5rN k= + = is odd, then Chebyshev differential matrix

rND is expressed as

rND =

-1.1708

2

-0.8944

0.6180

-2

-0.1708

1.618

-0.8944

0.8944

-1.6180

0.1708

2

-0.618

0.8944

-2

1.1708

neglect


( ) ( ), ,N Ni j N i N jD D

− −= −Symmetry property of Chebyshev differential matrix :

-1.1708

2

-0.8944

0.6180

-2

-0.1708

1.618

-0.8944

0.8944

-1.6180

0.1708

2

-0.618

0.8944

-2

1.1708

rND =

-1.1708

2

-0.8944

0.6180

-2

-0.1708

1.618

-0.8944

0.8944

-1.6180

0.1708

2

-0.618

0.8944

-2

1.1708

-1.1708

2

-2

-0.1708

Permute column by

r

TND P =

( )1, 2, 4,3P =

is symmetric

0.8944

-1.6180

-0.618

0.8944

1E =

2E =

( )1

21 2

3

4

uu

E Euu

⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠

is NOT symmetric

is faster ?

41.6

21.2859

-1.8472

0.7141

-0.9528

-8

-68.3607

-31.5331

7.3167

-1.9056

2.2111

17.5724

40.8276

12.6833

-10.0669

5.7889

-3.6944

-23.6393

-23.6393

-3.6944

5.7889

-10.0669

12.6833

40.8276

17.5724

2.2111

-1.9056

7.3167

-31.5331

-68.3607

-8

-0.9528

0.7141

-1.8472

21.2859

41.6

2r r rN N ND D D= ⋅ =

0r

0r

1r

1r

2r

2r

3r

3r

4r

4r

5r

5r

Redundancy in coordinate transformation [5]2 1 5rN k= + = is odd, then Chebyshev differential matrix 2

rND is expressed as

2rND =

-31.5331

7.3167

-1.9056

2.2111

12.6833

-10.0669

5.7889

-3.6944

-3.6944

5.7889

-10.0669

12.6833

2.2111

-1.9056

7.3167

-31.5331

neglect


( ) ( )2 2

, ,N Ni j N i N jD D

− −=Symmetry property of Chebyshev differential matrix :

2rND =

-31.5331

7.3167

-1.9056

2.2111

12.6833

-10.0669

5.7889

-3.6944

-3.6944

5.7889

-10.0669

12.6833

2.2111

-1.9056

7.3167

-31.5331-31.5331

7.3167

12.6833

-10.06691D = is NOT sym.

Permute column by ( )1, 2, 4,3P =-3.6944

5.7889

2.2111

-1.9056is NOT sym.

2D =

2r

TND P =

-31.5331

7.3167

-1.9056

2.2111

12.6833

-10.0669

5.7889

-3.6944

-3.6944

5.7889

-10.0669

12.6833

2.2111

-1.9056

7.3167

-31.5331

Row-major indexing: remove redundancy [1]

( ),ij i ju u r θ

( ) ( )( ),1 ,2 , , 1 , 2 ,, , , , , , ,T

i i i i m i m i m i Nu u u u u u uθ+ +

1 1ri N j Nθ≤ ≤≤ ≤ − 1Define active variable for and

total number of active variables is 2 6 12k Nθ⋅ = ⋅ =11

12

13

uuu

⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠

14

15

16

uuu

⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠

1

NOT ( )1 4 6 24rN Nθ− ⋅ = ⋅ = 2

31u =

θ

r

01 r=

1r

2r

3r

4r

51rNr r− = =

0θ 1θ 2θ 3θ 4θ 5θ 6θ0 π 2Nθ

θ π=

0r =

4

51 2 3 4 5 67 8 9 10 11 12

6

Index order21

22

23

uuu

⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠

24

25

26

uuu

⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠

78redundant 9

2u =10

1112

Index order

2 1rN k= + is odd, and


0 j k≤ ≤ , then 1 0k j k jr r− + ++ =

cos : 0i rr

ir i NNπ⎛ ⎞

= ≤ ≤⎜ ⎟⎝ ⎠

suppose since

( ) ( ) ( )2 11

1cos cos cosrN k

k j k jr r r

k j k j k jr r

N N Nπ π π

π = +− + +

⎛ ⎞ ⎛ ⎞− − + += = − − ⎯⎯⎯⎯→− = −⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

2 ,N mθ

π πθ∆ = = : 1j j j Nθθ θ= ∆ ≤ ≤2N mθ = is even, and

( )( )mod 2m j jθ π π θ+ + =Hence for andj m jθ π θ ++ =1 ,j m≤ ≤

0 1r =1rkr01kr +1rNr −1rNr− =

rk jr −1k jr + +

1 0k kr r ++ =

1 0k j k jr r− + ++ =

1 1 0rNr r− + =


( ) ( )( )( )1 1, , mod 2k i j k i ju r u rθ θ π π+ + + += − +From symmetry condition, we have

( ) ( )1 , ,k i j k i m ju r u rθ θ+ + − +=

0 i k≤ ≤for 0 j m≤ ≤ , symmetry condition impliesand( ) ( )1 , ,k i m j k i ju r u rθ θ+ + + −=

Therefore, we have two important relationships

1, , 1,

2, 1, 2,

1, 1, ,

11

1r

k j k m j m j

k j k m j m j

N j m j k m j

u u uu u u

u u u

+ + +

+ − + +

− + +

⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟= =⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠ ⎝ ⎠⎝ ⎠

1

1, , 1,

2, 1, 2,

1, 1, ,

11

1r

k m j k j j

k m j k j j

N m j j k j

u u uu u u

u u u

+ +

+ + −

− +

⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟= =⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠ ⎝ ⎠⎝ ⎠

2

θ

r

01 r=

1r

2r

31 r− =

00 θ= 1θ 2θ 3θ π=

Kronecker product [1]

1 2 3

4 5 6

11

12

13

uuu

⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠

21

22

23

uuu

⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠

1u =

2u =

1

2( ),ij i ju u r θ

1 2i≤ ≤ 1 3

Define active variable 3

4j≤ ≤andfor

56

Separation of variable: assume matrix A acts on r-dir and matrix B acts on dirθ −

( )( )( )

1 1,11 12

2,21 222

,

,

j j

jj

r ua aAu

ua ar

θ

θ

⎛ ⎞ ⎛ ⎞⎛ ⎞⎜ ⎟ = ⎜ ⎟⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠⎝ ⎠is independent of j

( )( )( )( )

1 11 12 13 ,1

2 21 22 23 ,

3 31 32 33 ,3

,,,

i i

i i

i i

r b b b uBu r b b b u

r b b b u

θθθ

⎛ ⎞⎛ ⎞ ⎛ ⎞⎜ ⎟⎜ ⎟ ⎜ ⎟= ⎜ ⎟⎜ ⎟ ⎜ ⎟

⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠ ⎝ ⎠

i

Let 1 6

2

uX R

u⎛ ⎞

= ∈⎜ ⎟⎝ ⎠

be row-major

index of active variable 2 is independent of


11 12 6 6

21 22

a B a BA B R

a B a B×⎛ ⎞

⊕ ≡ ∈⎜ ⎟⎝ ⎠

Kronecker product is defined by

( ) 11 12 1 11 1 12 2

21 22 2 21 1 22 2

a B a B u a Bu a BuA B X

a B a B u a Bu a Bu+⎛ ⎞⎛ ⎞ ⎛ ⎞

⊕ = =⎜ ⎟⎜ ⎟ ⎜ ⎟+⎝ ⎠⎝ ⎠ ⎝ ⎠

Case 1: A I=

( )( )( )( )

1 11 12 13 ,1

2 21 22 23 ,2

3 31 32 33 ,3

,,,

i i

i i

i i

r b b b uBu r b b b u

r b b b u

θθθ

⎛ ⎞⎛ ⎞ ⎛ ⎞⎜ ⎟⎜ ⎟ ⎜ ⎟= ⎜ ⎟⎜ ⎟ ⎜ ⎟

⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠ ⎝ ⎠

( ) 1

2

BuI B X

Bu⎛ ⎞

⊕ = ⎜ ⎟⎝ ⎠

Case 2: B I=

( )

11 21

11 12 12 22

13 2311 1 12 2

21 1 22 2 11 21

21 12 22 22

13 23

u ua u a u

u ua u a uA I X

a u a u u ua u a u

u u

⎛ ⎞⎛ ⎞ ⎛ ⎞⎜ ⎟⎜ ⎟ ⎜ ⎟+⎜ ⎟⎜ ⎟ ⎜ ⎟

⎜ ⎟ ⎜ ⎟⎜ ⎟+⎛ ⎞ ⎝ ⎠ ⎝ ⎠⎜ ⎟⊕ = =⎜ ⎟+ ⎜ ⎟⎛ ⎞ ⎛ ⎞⎝ ⎠⎜ ⎟ ⎜ ⎟⎜ ⎟+⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠

( )( )( )

1 1,11 12

2,21 221

,

,

j j

jj

r ua aAu

ua ar

θ

θ

⎛ ⎞ ⎛ ⎞⎛ ⎞⎜ ⎟ = ⎜ ⎟⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠⎝ ⎠


Case 3: permutation, if permute 2 3θ θ↔

θ

r

01 r=

1r

2r

31 r− =

00 θ= 1θ 2θ 3θ π=

1 2 3

4 5 6θ

r

01 r=

1r

2r

31 r− =

00 θ= 1θ 2θ π=3θ

1 2 3

4 5 6

11

12

13

uuu

⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠

21

22

23

uuu

⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠

11

13 1

12

uu P uu

θ

⎛ ⎞⎜ ⎟ = ⋅⎜ ⎟⎜ ⎟⎝ ⎠

21

23 2

22

uu P uu

θ

⎛ ⎞⎜ ⎟ = ⋅⎜ ⎟⎜ ⎟⎝ ⎠

( )1,3, 2Pθ = 1

2

uX

u⎛ ⎞

= =⎜ ⎟⎝ ⎠

1

2

uX

u⎛ ⎞

= =⎜ ⎟⎝ ⎠

( ) 1

2

P uI P X X

P uθ

θθ

⎛ ⎞⊕ = =⎜ ⎟

⎝ ⎠


Case 4: ( )1 2,A diag a a=

( ) 1 1

2 2

a BuA B X

a Bu⎛ ⎞

⊕ = ⎜ ⎟⎝ ⎠

( )( )( )( )

1 1 11 12 13 1,1

1 1 2 1 21 22 23 1,2

1 3 31 32 33 1,3

,,,

r b b b ua Bu r a b b b u

r b b b u

θθθ

⎛ ⎞⎛ ⎞ ⎛ ⎞⎜ ⎟⎜ ⎟ ⎜ ⎟= ⎜ ⎟⎜ ⎟ ⎜ ⎟

⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠ ⎝ ⎠

( )( )( )( )

2 1 11 12 13 2,1

2 2 2 2 21 22 23 2,2

2 3 31 32 33 2,3

,,,

r b b b ua Bu r a b b b u

r b b b u

θθθ

⎛ ⎞⎛ ⎞ ⎛ ⎞⎜ ⎟⎜ ⎟ ⎜ ⎟= ⎜ ⎟⎜ ⎟ ⎜ ⎟

⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠ ⎝ ⎠2

2

1 1 rr ru u u ur r θθ λ+ + = −

2,2 2 2

1 2

1 1 1, , , Nk

diag Dr r r θθ

⎛ ⎞⊕⎜ ⎟

⎝ ⎠

1i =

2i =

22

1 1rr ru u u u

r r θθ λ+ + = − , on [ ]1,1r∈ − [and ]0,2θ π∈

2 1rN k= + 2N mθ = is even, thenis odd, and

cos : 1ir

ir i kNπ⎛ ⎞

= ≤ ≤⎜ ⎟⎝ ⎠

: 1j j j Nθθ θ= ∆ ≤ ≤Active variable is

Total number is ,kNθ NOT ( )1rN Nθ− ⋅

Non-active variable active variable [1]

0, 0 2i jr, that is θ π> < ≤

2 22

1 1r rN N r ND u D u D u u

r r θλ+ + = −

x

y1θ2θ

3θ

4θ 5θ

6θ( )1 0u r = ± =

Note that differential matrix rND ( )1rN +is of dimension

0,

1,

1,

,

r

r

j

j

N j

N j

uu

u

u−

⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠

Neglect due to B.C.,acts on


2rND

rND

1,

2,

,

j

j

k j

uu

u

⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠

1,

2,

1,r

k j

k j

N j

uu

u

+

+

−

⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠

However and act on

NOT active variable, how to deal with?

From previous discussion, we have following relationships which can solve this problem

( )

1, , 1,

2, 1, 2,

1, 1, ,

: 1:1

r

k j k m j m j

k j k m j m j

N j m j k m j

u u uu u u

k

u u u

+ + +

+ − + +

− + +

⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟= = −⎜ ⎟ ⎜ ⎟ ⎜⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟

⎝ ⎠ ⎝ ⎠⎝ ⎠

( )

1, , 1,

2, 1, 2,

1, 1, ,

: 1:1

r

k m j k j j

k m j k j j

N m j j k j

u u uu u u

k

u u u

+ +

+ + −

− +

⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟= = −⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟

⎝ ⎠ ⎝ ⎠⎝ ⎠

⎟ and

( ): 1:1k −where is a permutation matrix


( ),ij i ju u r θ 1 1ri N j Nθ≤ ≤≤ ≤ − 1Recall for and

and evaluate at ( ),i jr θConsider Chebyshev differential matrix , rr ND acts on u

( )( ) ( ) ( ), ,, row of ,r rr N i j r N jD u r i D uθ θ= − ⋅ i

We write in matrix notation

( )( ) ( ), ,1 ,2 , , 1 , 2 , 1,r rr N i j i i i k i k i k i ND u r d d d d d dθ + + −=

0r > 0r <

1,

2,

,

j

j

k j

uu

u

⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠

1,

2,

1,r

k j

k j

N j

uu

u

+

+

−

⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠

0r >

0r <

Question: How about if we arrange equations on ( ), i jr iθ ∀ when fixed j


( )

( )( )

( )( )( )

( )

111 12 1, 1, 1 1, 2 1, 1

221 22 2, 2, 1 2, 2 2, 1

,1 ,2 , , 1 , 2 , 1

1,1 1,2 1,1

2

1

,

,

,

,

,

,

r

r

r

r

r

jk k k N

jk k k N

k j k k k k k k k k k NN

k k k kk j

k j

N j

rd d d d d d

r d d d d d d

r d d d d d dD u

d d d dr

r

r

θ

θ

θ

θ

θ

θ

+ + −

+ + −

+ + −

+ + ++

+

−

⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟

=⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠

1, 1 1, 2 1, 1

2,1 2,2

1,1 1,2 1, 1, 1 1, 1

r

r r r r r r

k k k k k N

k k

N N N k N k N N

d d

d d

d d d d d

+ + + + + −

+ +

− − − − + − −

⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠

1,

2,

,

j

j

k j

uu

u

⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠

1,

2,

1,r

k j

k j

N j

uu

u

+

+

−

⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠

,

1,

1,

k m j

k m j

m j

uu

u

+

− +

+

⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠

0r >

0r >

0r <

0r <

We only keep operations on active variable ( )0,u r θ>

( ), ju r θ

That is, only consider equation ( )( ),rN i jD u r θ for 0ir >

1 2

3 4rN

E ED

E E⎛ ⎞

= ⎜ ⎟⎝ ⎠

abbreviate

( )1 2rND E E=Later on, we use the same symbol


Define permutation matrix ( ) ( )( ) 11: , 1: 1: 1

1

r

I

P k N k

⎛ ⎞⎜ ⎟⎜ ⎟= − − + =⎜ ⎟⎜ ⎟⎝ ⎠

k

k

1,

2,

,

j

j

k j

uu

u

⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠

1,

2,

1,r

k j

k j

N j

uu

u

+

+

−

⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠

( )( ), jP u r θ

1,

2,

,

j

j

k j

uu

u

⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠

1,

2,

,

m j

m j

k m j

uu

u

+

+

+

⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠

,

1,

1,

k m j

k m j

m j

uu

u

+

− +

+

⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠

( ), ju r θ ≡Let

Active variable with the same indexing in r-direction


Moreover, we modify differential matrix according to permutation P by

( )( ) ( ) ( )( ), ,r r

TN j N jD u r D P Pu rθ θ=

( )11 12 1, 1, 1 1, 2 1, 1

21 22 2, 2, 1 2, 2 2, 11 2

,1 ,2 , , 1 , 2 , 1

r

r

r

r

k N k k

k N k kTN

k k k k k N k k k k

d d d d d d

d d d d d dD P E E

d d d d d d

− + +

− + +

− + +

⎛ ⎞⎜ ⎟⎜ ⎟= =⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠

0r > 0r <

where

such that for 1 j m≤ ≤

( )( )1, 1,

2, 2,, 1 2

, ,

,r

j m j

j m jr N j

k j k m j

u uu u

D u r E E

u u

θ

+

+

+

⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟= +⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

( )( )1, 1,

2, 2,, 1 2

, ,

,r

m j j

m j jr N m j

k m j k j

u uu u

D u r E E

u u

θ

+

++

+

⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟= +⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

and

( ) ( ) ( )( )1 2, , , , , ,j j k jr r rθ θ θEvaluated at


( ) ( ) ( ) ( )( )1,1 1,2 1, 1, 1 1, 2 1,2

2,1 2,2 2, 2, 1 2, 2 2,2, 1 2 1 2

,1 ,2 , , 1 , 2 ,2

, , , , , ,r

m m m m

m m m mr N m

k k k m k m k m k m

u u u u u uu u u u u u

D u r r r E E

u u u u u u

θ θ θ

+ +

+ +

+ +

⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟= +⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

( ) ( ) ( ) ( )( )1, 1 1, 2 1,2 1,1 1,2 1,

2, 1 2, 2 2,2 2,1 2,2 2,, 1 2 2 1 2

, 1 , 2 ,2 ,1 ,2 ,

, , , , , ,r

m m m m

m m m mr N m m m

k m k m k m k k k m

u u u u u uu u u u u u

D u r r r E E

u u u u u u

θ θ θ

+ +

+ ++ +

+ +

⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟= +⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

,1

,2

,

,

i

ii

i m

uu

U

u

⎛ ⎞⎜ ⎟⎜ ⎟≡ ⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠

, 1

, 2

,2

i m

i mi

i m

uu

V

u

+

+

⎛ ⎞⎜ ⎟⎜ ⎟≡ ⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠

( )1 1

2 21 2 1 2, ,

T T

T T

T Tk k

U VU V

X X X X X

U V

⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟= = =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

define and active variable

( ) ( ) ( )( ) ( ) ( ), 1 2 1 1 2 2 2 1, , , , | |rr N mD u r r E X X E X Xθ θ = +To sum up


( )1 1

2 21 2

T T

T T

T Tk k

U VU V

X X X

U V

⎛ ⎞⎜ ⎟⎜ ⎟= = ⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠( ) ( )1 1 2 2| | | | | | | | |

TT T T T T T T Tj j k kmem X U V U V U V U V=

Note that under row-major indexing, memory storage of

is

( ) ( )2 1 1 1 2 2| | | | | | | | | |TT T T T T T T T

j j k kmem X X V U V U V U V U=but

( ) ( )2 1| mk

m

Imem X X I mem X

I⎧ ⎫⎛ ⎞⎪ ⎪= ⊕ ⋅⎨ ⎬⎜ ⎟⎪ ⎪⎝ ⎠⎩ ⎭

If we adopt Kronecker products, then

( ) ( ) ( )( ) ( ), 1 2 1 2, , , ,r

m mr N m

m m

I ID u r r E E mem X

I Iθ θ

⎧ ⎫⎛ ⎞ ⎛ ⎞⎪ ⎪= ⊕ + ⊕⎨ ⎬⎜ ⎟ ⎜ ⎟⎪ ⎪⎝ ⎠ ⎝ ⎠⎩ ⎭

11 12 1

12 22 2

1 2

n

n

m m mn

a B a B a Ba B a B a B

A B

a B a B a B

⎛ ⎞⎜ ⎟⎜ ⎟⊕ =⎜ ⎟⎜ ⎟⎝ ⎠

Definition: Kronecker product is defined by


The same reason holds for second derivative operator

1 22,

3 4rr N

D DD

D D⎛ ⎞

= ⎜ ⎟⎝ ⎠

( ) ( ) ( )( ) ( )2, 1 2 1 2, , , ,

r

m mr N m

m m

I ID u r r D D mem X

I Iθ θ

⎧ ⎫⎛ ⎞ ⎛ ⎞⎪ ⎪= ⊕ + ⊕⎨ ⎬⎜ ⎟ ⎜ ⎟⎪ ⎪⎝ ⎠ ⎝ ⎠⎩ ⎭

( )2, 1 2rr ND D D=

0r

neglect equation on

<( )2

, 1 2r

Tr ND P D D=

then

( ) ( ) ( ), ,1 1, row of matrix

r r

Tr N i j r N

i

D u r i D Pr r

θ⎡ ⎤ = × −⎢ ⎥⎣ ⎦collect equations on each :1ir i k≤ ≤

1 21 m m

m m

I IR E E

I Ir r⎧ ⎫⎛ ⎞ ⎛ ⎞∂ ⎪ ⎪→ ⊕ + ⊕⎨ ⎬⎜ ⎟ ⎜ ⎟∂ ⎪ ⎪⎝ ⎠ ⎝ ⎠⎩ ⎭

we have

1

2

1 2

1/1/ 1 1 1, , ,

1/k

k

rr

R diagr r r

r

⎛ ⎞⎜ ⎟ ⎛ ⎞⎜ ⎟= = ⎜ ⎟⎜ ⎟ ⎝ ⎠⎜ ⎟⎝ ⎠

where


directionθ −We write second derivative operator on as

( ) ( )( )

( )( ) ( )

2

22

2

1 0 mod 6 3

1, 0 mod

2sin / 2

jN j

j Nh

S xj N

jh

π⎧− − =⎪⎪= ⎨ −⎪− ≠⎪⎩

( ) ( )( )22,N N k jD S x x

θθ ≡ − where

( )1 , 1 2 0ii k j m r≤ ≤ ≤ ≤ >for

( )( ),1

,22 2 2, , ,

,2

, row of row of

i

i iN i j N N

i

i m

uu U

D u r j D j DV

u

θ θ θθ θ θθ

⎧ ⎫⎛ ⎞⎪ ⎪⎜ ⎟ ⎧ ⎫⎛ ⎞⎪ ⎪ ⎪ ⎪⎜ ⎟= − = −⎨ ⎬ ⎨ ⎬⎜ ⎟⎜ ⎟ ⎪ ⎪⎝ ⎠⎩ ⎭⎪ ⎪⎜ ⎟⎜ ⎟⎪ ⎪⎝ ⎠⎩ ⎭

( )( )

,1

,22 2 2, , ,

,2

,

i

i iN i N N

i

i m

uu U

D u r D DV

u

θ θθ θ θθ

⎛ ⎞⎜ ⎟

⎛ ⎞⎜ ⎟= = ⎜ ⎟⎜ ⎟ ⎝ ⎠⎜ ⎟⎜ ⎟⎝ ⎠

( )2 2, ,2 2

1 1, iN i N

ii

UD u r D

Vr rθ θθ θθ⎛ ⎞⎛ ⎞ = ⎜ ⎟⎜ ⎟

⎝ ⎠ ⎝ ⎠so

θimplies


( )( )

( )

1 12 2,2 ,2

1 11 11

22 22,,2 22 2

2, 2 222

22 ,2,2

1 1

,11,1

,11

N N

NNN

kk

NNkkk

U UD DV Vr r

rU U

DDrrD u V Vr

rr

U DD rVr

θ θ

θθ

θ

θθ

θ θ

θθθ

θθ

θθ

θ

⎛ ⎞⎛ ⎞ ⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠⎜ ⎟ ⎜ ⎟⎛ ⎞ ⎜ ⎟ ⎜ ⎟⎛ ⎞ ⎛⎜ ⎟ ⎜ ⎟⎛ ⎞ ⎜ ⎟⎜ ⎟ ⎜⎜ ⎟ = =⎜ ⎟⎝ ⎠ ⎝⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎜ ⎟⎛ ⎞ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠⎝ ⎠

k

k

UV

⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟⎞⎜ ⎟⎟⎜ ⎟⎠⎜ ⎟⎜ ⎟⎜ ⎟⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

If we adopt Kronecker products, then

( )( )

( )

( ) ( )

1

22 2 2, ,2

,,1

,

N N

k

rr

D u R D mem Xr

r

θ θθ θ

θθ

θ

⎛ ⎞⎜ ⎟

⎛ ⎞⎜ ⎟ = ⊕⎜ ⎟⎜ ⎟⎝ ⎠⎜ ⎟⎜ ⎟⎝ ⎠


summary

( ) ( ) ( )( ) ( )2, 1 2 1 2, , , ,

r

m mr N m

m m

I ID u r r D D mem X

I Iθ θ

⎧ ⎫⎛ ⎞ ⎛ ⎞⎪ ⎪= ⊕ + ⊕⎨ ⎬⎜ ⎟ ⎜ ⎟⎪ ⎪⎝ ⎠ ⎝ ⎠⎩ ⎭

1

( ) ( )( ) ( ), 1 2 1 21 1 , , , ,

r

m mr N m

m m

I ID u r r R E E mem X

I Ir r rθ θ

⎧ ⎫⎛ ⎞ ⎛ ⎞∂ ⎪ ⎪⎛ ⎞→ = ⊕ + ⊕⎨ ⎬⎜ ⎟ ⎜ ⎟⎜ ⎟∂ ⎝ ⎠ ⎪ ⎪⎝ ⎠ ⎝ ⎠⎩ ⎭

( )( )

( )

( ) ( )

1

22 2 2, ,2

,,1

,

N N

k

rr

D u R D mem Xr

r

θ θθ θ

θθ

θ

⎛ ⎞⎜ ⎟

⎛ ⎞⎜ ⎟ = ⊕⎜ ⎟⎜ ⎟⎝ ⎠⎜ ⎟⎜ ⎟⎝ ⎠

2

3

Note that

( ) ( ) ( )( )( )( )

( )

( ) ( ) ( )( ) ( ) ( )

( ) ( ) ( )

1 1 1 1 2 1 2

2 2 1 2 2 2 21 2 2

1 2 2

, , , ,, , , ,

, , ,

, , , ,

m

mm

k k k k m

r r r rr r r r

r r r

r r r r

θ θ θ θθ θ θ θ

θ θ θ

θ θ θ θ

⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟= =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

so that all three system of equations are of the same order.


22

1 1rr ru u u

r r θθ[ ]1,1r∈ −u , on andλ+ + = − [ ]0,2θ π∈

Discretization on ( )mem X

( ) ( )2L mem X mem Xλ⋅ = −

( )

( ) ( ) ( )

1 2 1 2

2 2,

2 21 1 2 2 ,

m m m m

m m m m

N

m mN

m m

I I I IL D D R E E

I I I I

R D

I ID RE D RE R D

I I

θ

θ

θ

θ

⎧ ⎫ ⎧ ⎫⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎪ ⎪ ⎪ ⎪= ⊕ + ⊕ + ⊕ + ⊕⎨ ⎬ ⎨ ⎬⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎪ ⎪ ⎪ ⎪⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎩ ⎭ ⎩ ⎭

+ ⊕

⎛ ⎞ ⎛ ⎞= + ⊕ + + ⊕ + ⊕⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

then

1

2

1 2

1/1/ 1 1 1, , ,

1/k

k

rr

R diagr r r

r

⎛ ⎞⎜ ⎟ ⎛ ⎞⎜ ⎟= = ⎜ ⎟⎜ ⎟ ⎝ ⎠⎜ ⎟⎝ ⎠

where

Example: program 28

2u uλ∆ = − with boundary condition ( )1, 0u r θ= =

25rN = ( ),k kVλis even, let eigen-pair beis odd and 20Nθ =

( ) ( ) ( )2 21 1 2 2 ,

m mN

m m

I IL D RE D RE R D

I I θθ⎛ ⎞ ⎛ ⎞

= + ⊕ + + ⊕ + ⊕⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

sparse structure of

Dimension : 1 12 20 240

2rN Nθ−

= × =

Example: program 28 (mash plot of eigenvector)

1 Eigenvalue is sorted, monotone increasing and normalized to first eigenvalue

1 2 nλ λ λ≤ ≤ ≤ ≤

Eigenvector is normalized by supremum norm, kk

k

VVV

∞

←2

1 5.7832λ = 2 3 14.682λ λ= = 4 5 26.3746λ λ= = 6 30.4713λ = 7 8 40.7065λ λ= = 9 10 42.2185λ λ= =

Example: program 28 (nodal set)

Exercise 1

2u uλ∆ = − with boundary condition ( )1,2 0u r = =

25rN = is odd and 20Nθ = is even, let eigen-pair be ( ),k kVλ

{ }1 2r< <on annulus

{ }1 2r< <[ ]1,1x∈ −Chebyshev node on Chebyshev node on

112

xr += +

( ) ( )2 2 2, , ,r rr N r N N NL D RD I R D

θ θθ= + ⊕ + ⊕

Exercise 1 (mash plot of eigenvector)1 Eigenvalue is sorted, monotone increasing and normalized to first eigenvalue

1 2 nλ λ λ≤ ≤ ≤ ≤

Eigenvector is normalized by supremum norm, kk

k

VVV

∞

←2

Chapter 11 polar coordinate - National Tsing Hua...

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